First the weighted average for the molar mass of air.
(0.8 x molar mass N2) +(0.2 x molar mass O2) = about 28.8 but you can do that more accurately.
Next the pressure of the tire in psi. We need that in atmospheres. I went to google and typed in 46.7 psi to atm and it returned 3.178 atm.
Now use PV = nRT
You know P, you know V from the problem, you know R (0.08206) and you know T (24 C + 273 = ??K). Calculate n = number of moles air.
Then moles = grams/molar mass. yOu know moles (n) and you know molar mass air (about 28.8 g/mol), calculate grams. That's your answer.
We're doing gas problems in chemsitry. I have no idea where to even begin on this problem:
Air is 20% oxygen and 80% nitrogen. What is the mass of air in an automobile tire of 19.7 L and internal pressure of 46.7 PSI at 24 degrees celsius? (That pressure is the same as the 32 PSI difference you usually measure as the tire pressure 32 PSI + 14.7 PSI. You will have to use a weighted average for the molar mass of air.)
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